Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100100;
int n, nn, m, q, x, y, p, hd[N], lg[N], nt[N * 2], to[N * 2], fa[20][N], dep[N],
    a[20][N][12], ans[12], temp[12];
void add(int a[], int b[], int c[]) {
  int j = 1, k = 1;
  for (int i = 1; i <= 10; ++i)
    if (b[j] < c[k])
      a[i] = b[j++];
    else
      a[i] = c[k++];
}
void run(int u) {
  dep[u] = dep[fa[0][u]] + 1;
  for (int i = hd[u]; i; i = nt[i])
    if (to[i] != fa[0][u]) {
      fa[0][to[i]] = u;
      run(to[i]);
    }
}
void fly(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  int t = dep[x] - dep[y];
  for (int i = 0; i <= nn; ++i)
    if (t & (1 << i)) {
      memcpy(temp, ans, sizeof(ans));
      add(ans, a[i][x], temp);
      x = fa[i][x];
    }
  if (x == y) {
    memcpy(temp, ans, sizeof(ans));
    add(ans, a[0][x], temp);
    return;
  }
  for (int i = nn; i >= 0; i--)
    if (fa[i][x] != fa[i][y]) {
      memcpy(temp, ans, sizeof(ans));
      add(ans, a[i][x], temp);
      memcpy(temp, ans, sizeof(ans));
      add(ans, a[i][y], temp);
      x = fa[i][x];
      y = fa[i][y];
    }
  memcpy(temp, ans, sizeof(ans));
  add(ans, a[0][x], temp);
  memcpy(temp, ans, sizeof(ans));
  add(ans, a[0][y], temp);
  memcpy(temp, ans, sizeof(ans));
  add(ans, a[0][fa[0][x]], temp);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 2; i <= n; ++i) lg[i] = lg[i / 2] + 1;
  nn = lg[n] + 1;
  for (int i = 1; i <= n - 1; ++i) {
    scanf("%d%d", &x, &y);
    nt[i * 2 - 1] = hd[x];
    hd[x] = i * 2 - 1;
    to[i * 2 - 1] = y;
    nt[i * 2] = hd[y];
    hd[y] = i * 2;
    to[i * 2] = x;
  }
  run(1);
  memset(a, 127, sizeof(a));
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &x);
    int loc = lower_bound(a[0][x] + 1, a[0][x] + 11, i) - a[0][x];
    a[0][x][loc] = i;
  }
  for (int i = 1; i <= nn; ++i)
    for (int j = 1; j <= n; ++j) {
      fa[i][j] = fa[i - 1][fa[i - 1][j]];
      add(a[i][j], a[i - 1][j], a[i - 1][fa[i - 1][j]]);
    }
  while (q--) {
    int tot = 1;
    scanf("%d%d%d", &x, &y, &p);
    memset(ans, 127, sizeof(ans));
    fly(x, y);
    while (ans[tot] != 2139062143 && tot <= p) tot++;
    --tot;
    printf("%d", tot);
    for (int i = 1; i <= tot; ++i) printf(" %d", ans[i]);
    putchar('\n');
  }
  return 0;
}